Question

in triangle pqr, if qs is the angle bisector of angle q then show that A(triangle pqr)/A(triangle pqs)=pq/qr

Answer 1

in Δ pqr if qs bisect ∠q then Area of Triangle sqr / Area of Triangle pqs   = qr/pq

Step-by-step explanation:

Correct Question : A(triangle sqr)/A(triangle pqs)= qr/pq

Let say Angle q  = 2β

Then qs is bisector of ∠q

=> ∠pqs = ∠rqs = β

now Draw sm ⊥ pq    &  sn ⊥ qr

Area of Triangle pqs

= (1/2) * pq  * sm

in Δ smq

sinβ = sm/qs

=> sm = qs * sinβ

Area of Triangle pqs = (1/2) * pq  *qs * sinβ

Area of Triangle sqr

= (1/2) * qr  * sn

in Δ snq

sinβ = sn/qs

=> sn = qs * sinβ

Area of Triangle sqr = (1/2) * qr  *qs * sinβ

Area of Triangle sqr / Area of Triangle pqs   = (1/2) * qr  *qs * sinβ / (1/2) * pq  *qs * sinβ

=> Area of Triangle sqr / Area of Triangle pqs   = qr/pq